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As I thought I understood things, the Gamma matricies behave as the 4 orthogonal unit vectors of the Clifford algebra $\mathcal{Cl}_{1,3}(\mathbb C)$, (also the Pauli matricies are for the 3 of $\mathcal{Cl}_{3}(\mathbb C)$??).

But, I'm not getting results that are intuitive when I perform algebra using multivectors built out of these.

The dot and wedge products are $A\cdot B=\frac{1}{2}(AB+BA)$ and $A\wedge B=\frac{1}{2}(AB-BA)$, respectively. But, I should be able to wedge multiply all 4 unit vectors in order to achieve the unit pseudovector, but the result is zero when I attempt to use these definitions to matrix multiply $\gamma_0\wedge\gamma_1\wedge\gamma_2\wedge\gamma_3$.

I think I'm missing something large...

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2 Answers 2

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Those formulas for dot and wedge product are only valid when $A$ and $B$ are vectors, not if they are bivectors, trivectors, etc.

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    $\begingroup$ Moreover, because $\gamma_a \cdot \gamma_b = 0$ for $a \neq b$, $\gamma_a \wedge \gamma_b = \gamma_a \gamma_b$. This means $\gamma_0 \wedge \gamma_1 \wedge \gamma_2 \wedge \gamma_3 = \gamma_0 \gamma_1 \gamma_2 \gamma_3$. $\endgroup$
    – Muphrid
    Commented Jan 28, 2015 at 16:31
  • $\begingroup$ what does matrix multiplication with those yield then? is there no way to find the dot/wedge for blades/multivectors expressed as matricies in this way? $\endgroup$
    – Tyson
    Commented Jan 28, 2015 at 17:49
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    $\begingroup$ If the grades are known, you can derive the correct dot or wedge product formula, but I think it's easier to merely consider the geometric product (simple matrix multiplication in the matrix representation) and do a projection onto the subspace of multivectors of the proper grade (i.e. grade projection). $\endgroup$
    – Muphrid
    Commented Jan 28, 2015 at 23:38
  • $\begingroup$ can you point me to any resources which explain such a projection? $\endgroup$
    – Tyson
    Commented Feb 1, 2015 at 22:26
  • $\begingroup$ It seems that David Hestenes' site (geocalc.clas.asu.edu) isn't working anymore. Do you know where else I can read his works? $\endgroup$
    – mr_e_man
    Commented Jun 5 at 23:26
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You have to decide from the outset how many dimensions you want to work in.

In GA2 (2-D Geometric Algebra) A^B makes a 2-D bivector surface. There is no A^B^C that will "pop out" of the 2-D page, everything remains in 2-D.

In GA3 A^B^C makes a 3-D trivector volume, but you cannot "pop out" of 3-D into a fourth dimension, everything remains in 3-D.

And so on upward to higher dimensions.

It is part of the "closure" of GA, by the same principle that for a 1-D vectorA, A^A = zero and A.A = 1, so A*A = 1, collapsing a 1-D vector to a 0-D scalar.

Its like the "closure" of an odometer. Whenever it gets to 9999 +1 it rolls back around to 0000 to make sure your odometer does not "explode" if you exceed its limits, but simply rolls back to zero to start all over again, to make sure your calculation never "runs off the edge of the page" but remains within the dimensions of the algebra, like an "asteroids" spaceship that drifts off one edge of the screen only to reappear at the opposite edge.

It turns out that "closure" is one of the key features of GA seen in many different forms throughout the algebra.

Its explained here in simple intuitive terms.

https://slehar.wordpress.com/2014/03/18/clifford-algebra-a-visual-introduction/

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